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December 21, 2014

December 21, 2014

Posted by **Hannah** on Monday, March 7, 2011 at 3:14pm.

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Gas is escaping a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? ( A sphere of raius r has volume= 4/3 pir^3 and surface area S=4pir^2.)

Remember that ds/dt=ds/dr X dr/dt

Step 1: Find ds/dr

S=4pir^2 so ds/dr= (4pi)(2r) = 8pi r

Step 2: Fid dr/dt (Hint: dv/dt=dv/dr X dr/dt)

V=4/3pir^3 dv/dr=4pir^2 2=(4pir^2)(dr/dt) dr/dt=1/2pir^2

Step 3: Find ds/dt

ds/dt= (8pir)(1/2pir^2) = 4/r

Step 4: Evaluate ds/dt when the radius is 12ft.

r=12 (8)(3.14)(12) X (1/2(3.14)(12)^2) =4/12

(301.44)(1/904.32)=4/12

Are these correct?

- Math(Please check) -
**Reiny**, Monday, March 7, 2011 at 3:41pmcorrect, except reduce 4/12 to 1/3

I might have done it this way

V = (4/3)πr^3

dV/dt = 4πr^2 dr/dt , but when r=12, dV/dt = 2

2 = 4π(144) dr/dt

dr/dt = 1/(288π)

A = 4πr^2

dA/dt = 8πr dr/dt

= 8π(12)(1/(288π) = 1/3

- Math(Please check) -
**Hannah**, Monday, March 7, 2011 at 3:45pmOk thank you!

- Math(Please check) -
**Reiny**, Monday, March 7, 2011 at 4:09pmwelcome

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